The best way to answer this question is by examining each option provided as to its correctness regarding the concepts of organic chemistry involving carbanions, carbocations, and hybridization. Option A: The stability of carbanions is evaluated based on the ability of the surrounding groups to stabilize the negative charge. Generally, more aryl groups (such as phenyl groups, C6H5 ) around the negatively charged carbon increase the stability due to the delocalization of charge. Here, the correct stability trend should be (C6H5)3C−>(C6H5)2CH−>C6H5CH2−. This aligns with the assertion made in Option A, so Option A is correct. Option B: A methyl cation (CH3+)is not actually feasible under normal circumstances but assuming it exists in theory or in highly reactive environments, it is better described as sp2 hybridized. This setup consisting of three groups around the carbon atom assumes a trigonal planar arrangement rather than a tetrahedral (sp3) configuration. Thus, Option B is incorrect. Option C: In a methyl anion (CH3−), the negative charge is stabilized by distributing it over the additional lone pair which assumes a sp3 hybridization to maintain tetrahedral geometry, rather than sp2. So, this option is incorrect. Option D: The stability of carbocations generally increases with the number of alkyl groups attached to the positively charged carbon due to the inductive effect and resonance stabilization (if aryl groups are present). For aryl groups, they can help delocalize the positive charge through resonance stabilization. So, the order of stability is generally C6H5CH2+<(C6H5)2CH+<(C6H5)3C+, where more phenyl groups add greater stability. Therefore, Option D is also incorrect as it lists the stability order in reverse. In conclusion, the correct statement from the provided options is Option A.